SMD Indicator Bulbs

[bradams]

I was doing a late-night browse through the forum and came across a post from someone who had successfully fitted SMD/LED indicator bulbs. In the clear light of morning I cannot find the post! Anyone own up to it? :-)
I quite like the LED effect and would be interested in trying them. The only real problem is that I cannot fathom out how to access the rear bulbs! - secret-fixed plastic in front, body metal behind. How do you change the bulb?
(Also, next problem is - are decoy resistors required, in which case, which wires do I bridge?)

[SwissXC90]

Originally Posted by bradams

(Also, next problem is - are decoy resistors required, in which case, which wires do I bridge?)

First, I will ignore the legalities of what you are doing (which is in effect negating the type approval for your vehicle, and thus illegal)

LEDs use less power than normal bulbs

The car detects 2 x 21W and 1 x 5W indicator lamps as normal load.
(2 x 21) + 5 = 47W
Should a 21W bulb fail, then the load drops to 26W
And then the indicators flash twice as fast to tell you a bulb is out.

So you WILL need an effective resistor load of around 40-ish watts to fool the bulb failure detection. The remaining 7W will likely be the load of the LEDs.

Using Ohms law:
P = I x V and V = I x R
A 40W load at 14V is P/V amps = 40/14 = 2.85A
V = I x R
14V / 2.85A = 4.9 ohms

So you will need to fit around a 5 ohm resistor that can dissipate at lease 40W of heat.
Go for a 5 ohm 50W resistor
This will come in a metal body and will need to be mounted to metalwork to act as a heat sink to dissipate the heat.
Ensure there is no cloth or plastic in the area that could melt.

[bradams]

Originally Posted by SwissXC90

First, I will ignore the legalities of what you are doing (which is in effect negating the type approval for your vehicle, and thus illegal)

LEDs use less power than normal bulbs

The car detects 2 x 21W and 1 x 5W indicator lamps as normal load.
(2 x 21) + 5 = 47W
Should a 21W bulb fail, then the load drops to 26W
And then the indicators flash twice as fast to tell you a bulb is out.

So you WILL need an effective resistor load of around 40-ish watts to fool the bulb failure detection. The remaining 7W will likely be the load of the LEDs.

Using Ohms law:
P = I x V and V = I x R
A 40W load at 14V is P/V amps = 40/14 = 2.85A
V = I x R
14V / 2.85A = 4.9 ohms

So you will need to fit around a 5 ohm resistor that can dissipate at lease 40W of heat.
Go for a 5 ohm 50W resistor
This will come in a metal body and will need to be mounted to metalwork to act as a heat sink to dissipate the heat.
Ensure there is no cloth or plastic in the area that could melt.

Fascinating, I am sure, but I think we have moved on from the theory ......
All I want to know right now is how to get the b****y bulb out !!!

(and by the way, I already have a stock of 5ohm50w resistors; I was just wondering if this car design still requires them)